(square roots of the identity matrix) For how many 2x2 matrices A is it true that A^2=I ? Now answer the same question for n x n matrices where n>2

Step 1
To find how many matricec A of order 2 exists such that $A^2=I$
Here we have to find $A^2=A\times A=I$
Let us consider any arbitrary matrix of order 2. $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$
Then $A^2$ is given by $A^2=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}{a}^2+bc& ab+bd\\ ca+dc& d^2+bc\end{array}\right)$
Now from $A^2=I$ , we get $A^2=\left(\begin{array}{cc}{a}^2+bc& b(a+d)\\ c(a+d)& d^2+bc\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$
$(1)a^2+bc=1$
$(2)d^2+bc=1$
$(3)b(a+d)=0$
$(3)c(a+d)=0$
Step 2
Case 1:
If $(a+d)\ne 0$
From (3) we get : $b=0$
From (4) we get : $c=0$
From (1) we get : $a^2=1\Rightarrow a=\pm 1$
From (2) we get : $d^2=1\Rightarrow d=\pm 1$
Therefore we get the matrix of the form
$\{A=\left(\begin{array}{cc}a& 0\\ 0& d\end{array}\right):a=\pm 1,d=\pm 1\}$ Therefore there are only 4 matrices exist such that $A^2=I$ from case 1.
Step 3
Case 2:
If $(a+d)=0$
sub case a:
If a=0 , that implies $d=0$
Then from(1 ) and (2) , we get : $bc=1\Rightarrow c=b^{-1}=\frac{1}{b}$ in real numbers.
Therefore we get the matrix of the form
$\{A=\left(\begin{array}{cc}0& a\\ a^{-1}& 0\end{array}\right):a\in F\}$ Therefore there are infinitely many matrices exist if the field is the set of real numbers such that $A^2=I$
Step 4
We can also consider further subcases to find the type of involuntary matrices, but as the question is asked to find how many, we have already got infinitely many $2\times 2$ matrices of such type.
Similarly for any $n>2$ , we can find infinitely many matrices satisfying this condition on the field of real numbers.